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1:22 AM
@SirCumference pretty good - busy for sure =) how are you?
@EmilioPisanty this is my junior year, so i have two more years (including this one) before college
 
2:12 AM
@ACuriousMind I wrote an email to the person involved here:
https://physics.stackexchange.com/questions/502427/how-is-the-collapse-postulate-is-also-present-in-qft-only-hidden-inside-the-ls
He has replied along the lines of your answer :)
 
2:28 AM
@Rudi_Birnbaum Did I tell you the question you partially-answered (/commented) got answered!? All that's left is to assign a measure to L.H.S :)

https://math.stackexchange.com/questions/2888976/the-definite-integral-problem-with-a-twist
@heather I wish I was half as impressive as you! your a QC stackexchange mod too!
 
2:43 AM
@MoreAnonymous thank you...i don't think it's impressive, really, though. i just kind of bang my head against the wall until i figure something out.
 
@heather That's a great attitude to have! I remember I was quite the opposite in high school! I remember I would have a lot of the high school physics & math topics on my finger tips. Didn't really move on to more advanced topics. Though that's partially because in India your encouraged to focus on speed rather than insight of the subject.
 
 
4 hours later…
6:26 AM
@ScientistSmithYT I have no idea I'm afraid. I've never studied superconductivity in detail and know only the basics of it.
 
 
2 hours later…
7:59 AM
I'll just drop this link here (already shared it in QC chat):
 
Good morning!
Can I measure/calculate (sensically) the expectation value of an operator, say angular momentum, on a state eigenfunction from a Hamiltonian, that "violates the corresponding symmetry" in this case rotational symmetry. I think yes, the result is just any (real?) value depending on the nature of the state (One eg. could develop the eigenfunction in eigenfunctions of the momentum operator to work that out). Is that about right?
I got confused on it, since someone told me that would not be an "expectation value" in the actual sense.
@EmilioPisanty any take on that?
To be more specific lets take the angular momentum-$z$ component $\hat L_z$ and say $[\hat H,\hat L_z] \ne 0$. Then we want to calculate/interpret $<\psi |\hat L_z| \psi>$ for $\psi$ with $\hat H \psi = E \psi$.
 
8:26 AM
@Rudi_Birnbaum you're completely right
You can just calculate the expectation value
It will be real-valued whenever the observable is hermitian
 
@EmilioPisanty thank you very much! What troubles me is that the person is quite a "learned" scientist and that I do not get his point at all.
@EmilioPisanty OK. So for $\hat L_z$ its real in any case, right?
 
It generically won't coincide with any eigenvalue, so generically it won't match any individual experimental outcome
 
since $\hat L_z$ is hermitian "by itself".
@EmilioPisanty Yep I got that.
 
But it does coincide with the statistical expectation value as normally defined
 
Would there be any reasoning if and when it would get $=0$?
 
8:29 AM
@Rudi_Birnbaum no idea why they told you otherwise
 
@EmilioPisanty :-D ! Good!
 
@Rudi_Birnbaum could be any number of things
 
There is talk about "quenched angular momentum" ...
I mean in that case the spectrum of eigenvalues of its eigenfunctions is "symmetric around 0"
If all + eigenfunctions contribute equally to their - counterparts it would be the case
Just the question what that would mean
inversion symmetry possibly?
 
8:49 AM
@Rudi_Birnbaum I don't see any simple interpretation other than "equal weight on both sides of the spectrum".
I don't see how you'd extract anything useful about the wavefunction from that
The space of variations is too big and the amount of information isn't all that much.
 
@EmilioPisanty I saw an exercise in the net on <L^2> for the particle in the box its 0 and they argue the electron can travel equally to the left as to the right ..
L is dependent on the origin. So it kind of would make sense to conjecture certain symmetries
But I agree I cannot see the connection.
The relation between real space and (angular)momentum is Fourier Transform. So symmetries in the one should influence the other. But thats just vague heuristics. I also don't see way to prove that.
I mean $<L>=0 \Rightarrow$ is hopeless, but some nice $<L>=0 \Leftarrow$ might be doable.
 
@Rudi_Birnbaum eh
Interpreting angular momentum spectra as Fourier transforms is a stretch
 
@EmilioPisanty no interpretation intended, just an heuristics to an idea
 
heuristics is ok
 
8:58 AM
There's tons you can do with symmetry if it's an eigenstate with zero eigenvalue
But I don't see any connections to a zero expectation value
 
 
1 hour later…
10:13 AM
Huston Chairopractor here, bout ta break some booneees
 
10:46 AM
@EmilioPisanty Does the "first order density matrix" $\rho(x,x')=\int \Psi^* \Psi dx_2..dx_n$ for a n-electron Wavefunction $\Psi(x_1,..,x_n)$ obey a von-Neuman equation like "mixed state density matrices" in the one-particle case?
 
10:59 AM
And I mean for non-separable = entagled subsystems
s.t. $\Psi$ cannot be decomposed into a product.
 
11:53 AM
@Rudi_Birnbaum "first order density matrix" is an awful term
you only use "first order" if it's within a perturbation-theory framework
You probably mean the single-particle reduced density matrix
 
the thing given in the equation is what I mean
 
the equation isn't super clear
 
yeah it misses x and x' in the resecptive \Psi
Shall I write it again?
$\int\Psi^*(x,x_2,..x_n)\Psi(x',x_2,..x_n)dx_2 ... d x_n$
 
yeah I have googled that as well before I posted ...
 
11:58 AM
@Rudi_Birnbaum googled what?
 
The article you link
@EmilioPisanty this one
 
@Rudi_Birnbaum so there you have it
what else do you need?
 
understand the whole thing in order to be able to extract the answer to the question that for sure will only make a small but strongly entangeled part in the article.
 
I mean
the answer is
@Rudi_Birnbaum yes
 
The article seems to suggest it does not work with the one-particle RDM, but only with the
two particle
 
12:04 PM
whether it will work or not will depend on the system
@Rudi_Birnbaum depends on the system
sometimes the 1-RDM will work fine
sometimes you will need to use the 2-RDM
 
n electrons in the magnetic field
 
sometimes you'll need 3-RDMs or higher
 
in an nuclear force field
 
it depends on the configuration, the hamiltonian, and what observables you care about
 
I care about j
the probablitiy current
electronic ground state
 
12:05 PM
sometimes the whole RDM approach will be useless and fail to converge, sometimes it will be prohibitively expensive to calculate and you won't even be able to tell whether it converges
 
@EmilioPisanty Oh, interesting.
i think I will post a question.
 
To some this might be a scifi question rather than physics...what if the space between atoms increased and the "thing" still remained an atom? Does it mean more energy infused in the system?
 
@Rudi_Birnbaum This is nowhere near sufficiently detailed a description of the system to tell what methods will or won't work
 
for that its better to post a full question
the full is too long for the chat I suppose
 
@deostroll "what if physics were different? how would you analyze that situation within standard physics" is not a useful question, or one that we address here.
@Rudi_Birnbaum possibly
 
12:07 PM
What I actuall search for is a way to prove the continuity for the n-electron current
 
@Rudi_Birnbaum journals.aps.org/pra/abstract/10.1103/PhysRevA.95.033414 is probably clearer about the framework
@Rudi_Birnbaum that's a completely different game
 
Well my idea was to go via the 1P-RDM
 
7
A: QM Continuity Equation: Many-Body Version for Density Operator?

Mark Mitchison$\def\rr{{\bf r}} \def\ii{{\rm i}}$ There is indeed a continuity equation for the particle density $\rho(\rr)=\Psi^\dagger(\rr)\Psi(\rr),$ where the field operator $\Psi^\dagger(\rr)$ creates a particle at position $\rr$. To derive it, you need only the canonical commutation relations for the fie...

 
Hey! You're an ace!
 
12:10 PM
But is that for non-separable states ?
 
great!
 
I suspect you'll have the same reservations I did when I first read it
in that the interpretation of the whole thing is rather dubious surprising and counter-intuitive
but it works
 
@EmilioPisanty I know that since I do research on such currents for about 10 years ...
I was surprised when last week a paper came up that showed the continuity of such such effective one-particle currents
in -- to me -- very clumsy way
I always thought it holds more or less by definition ...
I post the link - wait
 
@Rudi_Birnbaum eh
all sorts of weird shit makes the publication bar
 
12:16 PM
What I would love to see, is a similar derivation like physics.stackexchange.com/questions/294837/… but using the 1-PDM instaed of second quantization if such a thing is possible ...
 
Yeah, it sounds reasonable
sounds like good material and good scope for a question
 
@EmilioPisanty in addition thats physics.stackexchange.com/questions/294837/… without magnetic field
I will try it, but thanks anyway!!
 
good luck =)
 
Thanks! :-)
 
1:06 PM
0
Q: QM Continuity Equation: many-electron in the magnetic field version?

Rudi_BirnbaumIn 1-particle non-relativistic QM we have the continuity equation for the 1-electron probability current density for an electron in the magnetic field in an eigenstate of the full Hamiltonian (=static problem) and ignoring spin $$ \nabla\cdot{\bf j}=0$$ with the definition of $$ {\bf j}=-\fra...

 
1:17 PM
@Rudi_Birnbaum ping me in two days if you don't have an answer, it definitely deserves some additional rep incentives
though maybe we can also ping @JohnRennie, who was complaining that he couldn't find good questions to bounty =P
 
1:38 PM
A guess: I think $A(r_i)$ is a 'one-particle' operator and on squaring the spin-less non-relativistic magnetic field kinetic term $(p - eA)^2/2m$ the $Ap$ and $A^2$ terms are 'two-particle operators', if they are non-separable the extra potential term $U$ in the Hamiltonian will split into a sum of 1 + 2+ .. + n particle operators which can also be expressed in second quantized form, then you would do what that post does just deriving $\rho$
Here is a plot of some resonances of the $\Delta^+$ baryons
So $\Delta_{3/2^+}$ of mass 1232 is the one commonly discussed, what's going on with e.g. $\Delta_{11/2^+}(2420)$? It's an 'excited state' of the 1232, but what does that mean - usually $\Delta^+(1232)$ is discussed as arising from a collision of $\pi^+$ with protons, I'm guessing the larger the speed of the initial $\pi^+$ the larger the mass of the $\Delta$ produced, is that even right, and how is the spin changing up to $11/2$...
 
2:00 PM
Here's a random question: Given in the lab I can measure $x$ and $p$ why can't I measure $xp + px$? Like I mean all the operators are hermitian? Why does nature exclude $xp + px$ from the possible operators that can be measured in a lab?

Where x is the position operator and p is the momentum operator?
Is there something obvious I don't know? ^
 
@EmilioPisanty Yes I will, thanks :)
@MoreAnonymous xp + px is not Hermitian, if xou could prove that you would have proved the RH.
@MoreAnonymous it seems to be a closely related operator with the same possible spectrum could be PT symmertric
 
2:16 PM
@Rudi_Birnbaum You sure? $(px)^{\dagger}= x^\dagger p^\dagger = xp $
 
and PT symmetry would also be connected with real eigenvalues
because hermiticity is not an exact condition
@MoreAnonymous Ill send you a link to a mse discussion
 
@Rudi_Birnbaum Can we do standard QM before we move onto PT (also PT Hamiltonian have equivalent standard QM Hamiltonian)
 
sure, since px is a monster!!
@MoreAnonymous parity-time
 
or maybe it was cp i have to check
 
2:20 PM
"By a nonunitary similarity transformation, the PT-symmetric Hamiltonian becomes hermitian. This corresponds to a change of the inner product. Thus nature is quantum mechanical, and in a few cases, this can be expressed in terms of a noncanonical inner product and a corrresponding PT-symmetric Hamiltonian. Nothing fundamentally new is going on."
@Rudi_Birnbaum Can we please stick to standard QM?
 
@MoreAnonymous ypu have started with px, nothing os standard with an operator like that
 
In mechanics, the virial theorem provides a general equation that relates the average over time of the total kinetic energy, ⟨T⟩, of a stable system consisting of N particles, bound by potential forces, with that of the total potential energy, ⟨VTOT⟩, where angle brackets represent the average over time of the enclosed quantity. Mathematically, the theorem states ⟨ T ⟩ = − 1 2 ∑ k =...
@Rudi_Birnbaum Remember him^ ?
I know the quantum version of the scalar $\hat G \neq px + xp$ but isn't intuitive on the classical level
 
search "riemann bender" on mse and then the one with the most votes
the answer is from bellisard himself
 
@Rudi_Birnbaum what?
 
@EmilioPisanty Well on $\mathcal{H}$ not on $L^2$
 
2:32 PM
what the hell is $\mathcal H$?
 
@Rudi_Birnbaum Can we please stick to the question I ask :( without PT. We'll talk about PT symmetric another day)
 
@MoreAnonymous what do you think PT means?
 
@Rudi_Birnbaum if any of the discussion regarding the hermiticity of $xp+px$ above depends on the distinction between hermitian / symmetric / self-adjoint, or on the space, or any other such technical matters, then please mind your audience and be very explicit about all those technical concerns.
 
@EmilioPisanty OK!
 
@MoreAnonymous for all intents and purposes as far as physics is concerned, within standard QM (and possibly requiring a physicist's cavalier attitude towards the precise details of the underlying functional analysis), $xp+px$ is hermitian.
 
2:35 PM
@EmilioPisanty Yes, my question is why isn't there any measuring apparatus for such a hermitian operator? Like is there some way to detect is in the lab?
 
Sorry I assumed @MoreAnonymous was talking about something else!
 
@MoreAnonymous that depends on the specific implementation.
 
$xp + px$ is equal to (say) $2xp + i \hbar$ and so it's expected value is just a scaled shifted version of the expected value of $xp$ i.e. the expected result of measuring first momentum then position
 
@bolbteppa sorry, but that's dead wrong
> the expected result of measuring first momentum then position
that has nothing to do with $\hat x\hat p$
 
@bolbteppa Isn't xp anti-hermitian -why should it matter?
@bolbteppa Also don't follow for the reason Emilo mentions :/
 
2:41 PM
@EmilioPisanty $\mathcal{H} = L^2(0,\infty)$
 
@EmilioPisanty The only implementation I think it can be is when your Hamiltonian is $xp +px$. Is there another non-trivial version?
And don't ask me what the kinetic energy is ^
 
@Rudi_Birnbaum I don't think it's appropriate to switch to such a pathological space without making that very clear, particularly when speaking with non-experts
@MoreAnonymous $xp+px$ is not suitable as a hamiltonian, because it's not bounded from below
by "implementation" I mean the specific details of what system you've chosen to instantiate your harmonic oscillator
i.e. it could be a mechanical oscillator, the vibrations of an atom in an atom trap, the field quadratures of a mode of electromagnetic radiation, etc etc etc etc etc etc etc
there are millions of systems governed by the 1D Schrödinger dynamics
 
@EmilioPisanty @MoreAnonymous was talinkg previously about the Bender work on the QM attack of the Riemann hypotheses, and there its the only and essential Hilbert space that counts. It turned out that the Physcsit Bender missed that. Its a very famous story and that operator plays the fundament role there
 
the experimental capabilities for each of them will be different.
 
@EmilioPisanty Wait so what's the name of the instrument that measures this eigenket?
The eigenket of $px + xp$
?
 
2:45 PM
@MoreAnonymous the idea that there is one unique such instrument is wrong
again
it depends on the platform
sorry, I'm at work and I don't have time for discussions that go in circles
3
 
@EmilioPisanty That's fine. Maybe i'll re-read your messages and google a bit to understand what your going on about
 
@EmilioPisanty in part the story got soo famous since Bellissard a well-known mathematician gave his full valid critics for the first time on MSE, namely here
Since then Bender and co reacted to the critics and now the focus is on the question if this or a similar Hamiltonian might be PT symmetric. (just saying). Sorry again!
on $L^2(0,\infty)$ ...
 
@EmilioPisanty that is ridiculous
 
3:34 PM
wot
r u ok bro
 
sorry, wrong chatroom, never mind XD :)
 
did u have a stroke?
ok
 
@enumaris probably, XD
sup bro?
 
nm
my GCP instance won't start
says they are out of resources...
 
@enumaris Code of Good practices? GCP?
 
3:36 PM
google cloud platform
 
@enumaris show the screenshot
 
can't really screenshot it cus it's a corporate account
 
@enumaris works for me... I've checked now !
 
but it says they are out of resources
in the central region
 
@enumaris what your country/region, they differ from region to region
 
3:38 PM
us-central
 
@enumaris can't say, works in India
 
I'll see if I can call them lol
 
@enumaris how it happened?
@enumaris try amazon AWS, they have better services
 
I can't move everything over lol it's a corporate thing
 
oh okay, leave it for a moment to see if things get fixed after 24 hours (if it's urgent) then put your head on a call and arrange some alternatives for ti
*it
@enumaris hey, I'm making a new Neural Net which can differentiate between music and noise and can separate noise out of music
 
3:44 PM
nice
denoising technology is pretty neat :)
 
yeeeeh!
 
there's a start-up working on it called Babblelabs
you can look into them if you are interested
 
@enumaris yes, I'll be writing a paper into Arxiv if it gets over
@enumaris I'm thinking to start my own startup
 
nice
I saw Babblelabs presentation at the last Nvidia conference
it seemed to have worked quite well
though it was general de-noising
not specifically for music
 
oh okay, mine will work in realtime
I'll also be working on it further to make an AI to generate Music itself with NLP to write lyrics itself then use autotune with it. next, Expert systems to write structure (like Chorus/hooks/raps/bridge etc.) Goal is to get everything in realtime
 
4:03 PM
nice
 
4:28 PM
@heather Life's been good, can't complain.
@heather Whoa
I'm starting to feel that whole "time speeds up when you get older" thing...last I recall you were in middle school and I was starting college
Oh god it's already 2020
 
now what
 
...I gotta chill
 
@SirCumference I too sometimes feel like I'm getting grey and old
 
I'm pretty sure I'll lose my hair before it grows grey
 
no energy left, want to live settled and die peacefully... XD
 
4:39 PM
@AbhasKumarSinha You're in high school tho :P
 
@SirCumference I'm fully exhausted tho
 
For me the whole "time speeding up" thing started in sophomore year of college and it's been super weird
 
Hi, anybody know the exact reason why adding less than optimal level water but just enough water to drench and dive the whole sasuage in during boiling in a pot isn't the optimal quicking cooking method? My intiuion is that adding a swimming pool of water or a few drip of water won't cook the sasuage.
 
@SirCumference and it will get worse
 
@SirCumference I can't even bear collage, will save money and will travel and live peacefully, It feels like I'm getting older and older day by day
 
4:40 PM
@Loong So I've heard
Like jeez, I've already lived more than a quarter of my life
 
I've probably lived about 1/4th of life...
 
@SirCumference Plenty of time left for immortality to be discovered
 
Also weird is that the 2010s are about to end. Soon we'll be living in the "20s" and we'll hear kids say "I grew up in the 20s"
 
No one dies a virgin life f***s everyone XD
@SirCumference never thought that way before
 
Hi, anybody here know some physics and can answer my question, thank you!
 
4:42 PM
@AbhasKumarSinha Yep, less than 90 days until then
 
@ACuriousMind Hope that doesn't get discovered too eartly
yes
 
@AbhasKumarSinha What, you want to become immortal as a frail old man? :P
 
@ACuriousMind It feels I'm a fragile old man but I'm actually a high school student who is burdened with a lot of pressure of studies, future, family, expectations and is getting exhausted
 
@ACuriousMind that hit's close to home...my hair is already thinning T__T
 
@enumaris mine too, I can feel that and you can see it
X_X
 
4:45 PM
Ugh early hair loss runs in my family
 
XD
@SirCumference you can try hair plantations, here it's cheap
 
I think my hair thinned out a lot in like a 2 year period and now it's kind of...stableized..
 
@AbhasKumarSinha I don't have any hair loss (yet?), so I'll put it off of my mind
 
welp, I was totally unable to get my instance up and running
 
@SirCumference That's good, but hair loss isn't something to worry about, loss of happiness hits worse.
 
4:47 PM
I'm getting hangry
and I have two meetings coming up...
better eat some beef jerky
 
@AbhasKumarSinha Yeah I know
 
Lately I've been hard to reach
I've been too long on my own
Everybody has a private world
Where they can be alone
Are you calling me, are you trying to get through
Are you reaching out for me, and I'm reaching out for you
I'm just so f**kin' depressed
I just can't seem to get out this slump
If I could just get over this hump
But I need something to pull me out this dump
I took my bruises, took my lumps
Fell down and I got right back up
But I need that spark to get psyched back up
In order for me to pick that mic back up
 
@AbhasKumarSinha You can become happy after being unhappy. You can't become hairy after getting unhairy!
 
@ACuriousMind That's not true, no one in my family has faced hairloss before
 
4:49 PM
I'm waiting for some gene therapy or something to come along
save me from my baldness
 
@ACuriousMind Hair plantations cost 1$ USD for 70 hairs, full plantations can be done in less than 7k USD
@enumaris Come to India, cheap hair plantations
 
yeah @ACuriousMind how are we gonna get immortality if we can't even perfect hair therapy
 
I feel like hair transplants is really hit or miss...they can look uber fake
then it's even worse than being bald
 
@enumaris "We'll splice some animal DNA into yours you make you grow back a luscious mane. Benefits: Hair loss is reverted. Risks: You might occasionally meow."
 
an acceptable risk
 
4:51 PM
@enumaris no, some good quality plantations have very natural like hair,
 
@SirCumference We'll just be immortal and bald.
 
@ACuriousMind that's how spiderman was made
 
yeah the good quality ones maybe...but who knows...
 
@ACuriousMind don't worry, hair loss is not a big problem, very less people have that
@enumaris no, they'll show you the catalog before doin' that.
 
well how about we lighten the mood with some memes
 
4:53 PM
@SirCumference hehehe
 
the BDFL recently stepped down
now he's former-BDFL
or maybe...BDFL-emeritus
 
@SirCumference And what are you gonna do with it while half the libraries you want to use are still on Python2? :P
 
oh it happened last year...
I only recently learned of it lol
python 2 will be deprecated end of this year I think
 
@enumaris Deprecation doesn't make it go away
 
@enumaris wait what
whoa it's true
man first Flash and now this. everything's going the way of the dodo in 2020
 
4:57 PM
It should
 
I found an alien that actually exists, biggest research for humanity in this century: images-na.ssl-images-amazon.com/images/I/…
Does anyone care about next-generation language? Man python is old and grey XD :P :)
 
@ACuriousMind but maybe if you give all the python2 coders a stern look...
 
@enumaris Plenty of unmaintained Python2 code lying around still!
If your project depends on that for some reason, you're stuck
 
lets talk about assembly language, it's pure ass!
 
you can port the libraries over to python 3
 
5:01 PM
@SirCumference In contrast to the dodo, Flash deserved it :P
 
@ACuriousMind Eh am I the only one who has fond memories of playing flash games in the 2000s?
like those were everywhere
 
@SirCumference That you could build nice stuff with it doesn't mean the thing itself was a good idea.
 
how am i going to play on coolmathgames.com without flash?
 
@SirCumference i too, I remember buying magzines to get free CD(s) which had 100(s) of Flash games, spongebob, horror craft, spiderman, scoobydoo, doraemon, courage and cowardy dog etc.. I used to complete my studies early just to take a bath before 9 and go to play after that.
 
@AbhasKumarSinha Yooo I remember getting those CD games
If we're talking about the same thing that is
 
5:05 PM
@SirCumference hehehe... Those memories, that's for reminding I missed those days/
 
@ACuriousMind Still though, it's just odd that it actually died out. Flash used to be everywhere
 
@SirCumference I too think/
@SirCumference flash isn't a software, it's emotions....
b4n, will join after taking dinner :) XD
Thanks, good day btw
@SirCumference Thanks sir for reviving those old days, I'm feeling better
 
Lol np
 
@EmilioPisanty Alright! I think I vindicated my intuition? Let's look at the eigenfunctions of $xp +px$. We write the eigenvalue equation as: $ x \frac{\partial \psi}{\partial x}+ \frac{\partial (x\psi)}{\partial x }= \lambda \psi$ where $\lambda$ is eigenvalue. The eigen-function $\psi=\lambda A$ where A is a constant. Obviously should not be an allowed state. Like what is the normalization? How does the Born rule work out $\int \phi^* \psi dx$ ? Hence, $xp + px$ is not an observable.
where $\phi^*$ is another wave function
@Rudi_Birnbaum you might me interested too ^
 
5:22 PM
@MoreAnonymous Your eigenfunctions are wrong. Did you even try to plug your solution back in? (It gives $\lambda A = \lambda^2 A$, which is obviously wrong for $\lambda \neq 0,1$) The solutions to your equation are $\psi(x) \propto x^{(\lambda - 1)/ 2}$.(for $\lambda = 1$ this is the case of your constant function)
 
@ACuriousMind Sorry made a silly mistake really running out of paper hear. I'll delete that post?
 
@JohnRennie Oh ok, well what do you know about it? Maybe you know something I don't know yet.
 
@MoreAnonymous Also, "the eigenfunctions are not normalizable" is not an argument to say something is not an observable. The eigenfunctions of $x$ and $p$ alone are plane waves/delta functions and aren't normalizable, either!
Non-normalizable eigenfunctions typically just mean that your operator has unbounded and continuous spectrum in at least one direction and that its eigenfunctions are not physically realizable states. Superpositions of them can still be perfectly fine, normalizable, states.
 
@ACuriousMind I see.
 
Hilbert spaces
There's a good overview in Balentine
 
5:34 PM
(I'm also wondering how you ran out of paper solving an equation I did in my head, but that's beside the point :P)
 
paper shows your errors easier
 
@ACuriousMind I was brute forcing ideas :/ Now I'm gonna have to do it slowly. Type on latex to calculate :'(
 
I don't know what any of that means
 
$\LaTeX$
you have to capitalize correctly to get it to show...ok...
 
@ACuriousMind It means you've tried various starting points on the same problem and none of them panned out :/ (plus there were few pages to being with)
 
5:43 PM
That equation is just $2x\psi'(x) = (\lambda - 1)\psi(x)$. Since $xf'(x) = nf(x)$ for any $f(x) = x^n$ ($n\neq 0$), my solution follows.
@enumaris Did you expect something designed by typography nerds not to care about capitalization? :P
 
well since \latex doesn't work it's not like it would overload the operator
 
6:11 PM
@ACuriousMind in a world with say 3 times and 3 dimensions, wouldn't the Dirac equation no longer be enough since the special orthogonal group connected to the identity is no longer simply connected
In mathematics, the indefinite orthogonal group, O(p, q) is the Lie group of all linear transformations of an n-dimensional real vector space that leave invariant a nondegenerate, symmetric bilinear form of signature (p, q), where n = p + q. The dimension of the group is n(n − 1)/2. The indefinite special orthogonal group, SO(p, q) is the subgroup of O(p, q) consisting of all elements with determinant 1. Unlike in the definite case, SO(p, q) is not connected – it has 2 components – and there are two additional finite index subgroups, namely the connected SO+(p, q) and O+(p, q), which has 2 components...
 
@bolbteppa "enough" for what?
 
To describe reality - why does the Dirac equation arise? Because $SO^+(1,3)$ is doubly-connected (plus details, but for three times this part changes)
 
If your question is if there are more general "spinors" in signature (3,3) than those induced by the Clifford algebra trick that gives us the Dirac representation, then yes
 
To what age is your brain more easily "ensmartened"? ... I don't know how to ask the question
I know that childhood and early adolescence are the best period to read books and "become smart" since that's the period during the brain is most plastic.
 
Right, Clifford algebras provide a way to represent the double cover, but if we lived in a world with 3 times the Dirac trick would not have given the most fundamental reps of the whole Lorentz group
and even worse a lot of what goes on in QFT books with SL(2,C) and Weyl spinors is special to 4D
 
6:20 PM
Until what age is the brain most plastic?
 
Alright, simply asked, have I missed the opportunity to improve my cognitive skills? I'm 17, can I still grow cognitively?
@bolbteppa Thanks
@Loong Thanks
I often feel like I should have read a lot when I was younger and I've missed the opportunity since my brain plasticity is lower now :(
I'm 17, do I still have time and hope to become as cognitively intelligent as possible by reading and learning a lot?
 
I don't know what "cognitively intelligent" means.
 
@ACuriousMind I knew you would say that. IQ? The ability to spot patterns? "Being Smart"?
I honestly don't know what "being smart" means nor do I think IQ makes any sense.
 
6:28 PM
So what are you worrying about? If you want to learn things, go ahead and learn them!
If you don't, don't. But why worry about some nebulous idea of "being smart" if you can acquire well-defined skills and knowledge instead?
 
Exactly the words I needed to hear (or read :D) :)
If "smartness" isn't properly defined, why do we call some people smarter and other dumber? Why kids who read early in life grow up to become geniuses?
 
@NovaliumCompany Many words we use in colloquial speech are not "properly defined"
And even then, we certainly recognize different kinds of smart, cf. "street smart" vs "book smart".
 
@ACuriousMind Exactly. Thank you! I'm just seeing those entrepreneurially successful people with high IQs and I'm like... will I be good enough, is there time for improvement?...
 
You would be surprised how many "high-IQ" people there are out there that are not "successful".
Being reasonably intelligent (in whatever nebulous sense) is maybe a necessary condition for success (though I doubt even that), but by no means sufficient. The idea that smart, hard-working people are all successful and all dumb, lazy ones are not is a pernicious myth.
 
@ACuriousMind Got it. Thanks for the therapy session :P
 
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